Answer-
The equations of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit are
[tex]12x-5y+14=0 \\ 12x-5y-14=0[/tex]
Solution-
Let a point which is 1 unit away from the line 12x-5y-1=0 is (h, k)
The applying the distance formula,
[tex]\Rightarrow \left | \frac{12h-5k-1}{\sqrt{12^2+5^2}} \right |=1[/tex]
[tex]\Rightarrow \left | \frac{12h-5k-1}{\sqrt{169}} \right |=1[/tex]
[tex]\Rightarrow \left | \frac{12h-5k-1}{13} \right |=1[/tex]
[tex]\Rightarrow 12h-5k-1=\pm 13[/tex]
[tex]\Rightarrow 12h-5k=\pm 14[/tex]
[tex]\Rightarrow 12h-5k=14,\ 12h-5k=-14[/tex]
[tex]\Rightarrow 12h-5k-14=0,\ 12h-5k+14=0[/tex]
[tex]\Rightarrow 12x-5y-14=0,\ 12x-5y+14=0[/tex]
Two equations are formed because one will be upper from the the given line and other will be below it.
